Step 1: Understanding the Substitution Process
When 10% of a mixture is replaced with pure adulterant (water), the mixture's concentration reduces to 90% of its original concentration.
Generally, if a proportion \( p \) of a mixture is substituted with pure adulterant, the new concentration becomes \( (1 - p) \) times the preceding concentration. In this case, \( p = \frac{4}{40} = 0.1 \).
The initial mixture is pure milk, with a concentration of 100% or 1. After \( n \) substitution cycles, the milk concentration is \( 1 \times (0.9)^n \).
Step 2: Finding the Smallest \( n \) for Milk Concentration Less Than 50%
We need to find the minimum number of substitutions, \( n \), such that the milk concentration falls below 50%. This can be written as:
\( 1 \times (0.9)^n<0.5 \).
Step 3: Solving the Inequality
Applying logarithms to both sides yields:
\(\log \left( (0.9)^n \right)<\log (0.5)\)
Using logarithmic properties, this simplifies to:
\(n \log (0.9)<\log (0.5)\)
Step 4: Calculating the Logarithms
Approximate logarithm values are:
\(\log (0.9) \approx -0.045757\) and \(\log (0.5) \approx -0.3010\)
Substituting these values into the inequality:
\(n \times (-0.045757)<-0.3010\)
Solving for \( n \):
\(n>\frac{-0.3010}{-0.045757} \approx 6.58\)
Step 5: Conclusion
The smallest integer greater than 6.58 is \( n = 7 \).
The minimum number of substitutions, \( n \), that meets this criterion is 7.
The correct option is (B): 7.